3. let ( f ) be a function defined by ( f(x)=1 - 2sin x;xleq0 ) ( =e^{-4x};x > 0 ) show that ( f ) is…
3. let ( f ) be a function defined by ( f(x)=1 - 2sin x;xleq0 ) ( =e^{-4x};x > 0 ) show that ( f ) is continuous at ( x = 0 ) 4. let ( f ) be the function given by ( f(x)=\frac{x}{sqrt{x^{2}-4}} ). (a) find the domain of ( f ). (b) write an equation for each vertical asymptote to the graph of ( f ). (c) write an equation for each horizontal asymptote to the graph of ( f ). (d) find ( f^{prime}(x) ).
Answer
Explanation:
Step1: Find the left - hand limit
For (x\leq0), (f(x)=1 - 2\sin x). (\lim_{x\rightarrow0^{-}}f(x)=\lim_{x\rightarrow0^{-}}(1 - 2\sin x)) Using the property (\lim_{x\rightarrow a}(u(x)-v(x))=\lim_{x\rightarrow a}u(x)-\lim_{x\rightarrow a}v(x)) and (\lim_{x\rightarrow0}\sin x = 0) (\lim_{x\rightarrow0^{-}}(1 - 2\sin x)=1-2\lim_{x\rightarrow0^{-}}\sin x=1 - 2\times0 = 1)
Step2: Find the right - hand limit
For (x > 0), (f(x)=e^{-4x}). (\lim_{x\rightarrow0^{+}}f(x)=\lim_{x\rightarrow0^{+}}e^{-4x}) Let (t=-4x), when (x\rightarrow0^{+}), (t\rightarrow0). (\lim_{x\rightarrow0^{+}}e^{-4x}=\lim_{t\rightarrow0}e^{t}=e^{0}=1)
Step3: Find the value of the function at (x = 0)
For (x = 0), using (f(x)=1 - 2\sin x) (since (x = 0) satisfies (x\leq0)) (f(0)=1-2\sin(0)=1)
Step4: Check the continuity condition
A function (y = f(x)) is continuous at (x=a) if (\lim_{x\rightarrow a^{-}}f(x)=\lim_{x\rightarrow a^{+}}f(x)=f(a)) Here, (\lim_{x\rightarrow0^{-}}f(x)=\lim_{x\rightarrow0^{+}}f(x)=f(0) = 1)
Answer:
Since (\lim_{x\rightarrow0^{-}}f(x)=\lim_{x\rightarrow0^{+}}f(x)=f(0) = 1), the function (f(x)) is continuous at (x = 0)